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Math Help - Find all real solutions

  1. #1
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    Find all real solutions

    1. 4x^4 - 18x^2 = 0

    2. \sqrt[3]{3x + 1} - 5 = 0

    it says to find all real solutions for the 1st problem, and find all solutions for the second problem...then it says to check

    I was absent today, but managed to get the homework. Can someone please explain how to do these?
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  2. #2
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    Quote Originally Posted by isundae View Post
    1. 4x^4 - 18x^2 = 0

    2. \sqrt[3]{3x + 1} - 5 = 0

    it says to find all real solutions for the 1st problem, and find all solutions for the second problem...then it says to check

    I was absent today, but managed to get the homework. Can someone please explain how to do these?
    1. Start by factorising. Then use the null factor law.

    2. Add 5 to both sides and then cube both sides. Make x the subject.
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  3. #3
    Senior Member pacman's Avatar
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    1.

    2.

    Solution:

    1)

    Use factoring,

    x^2(4x^2 - 18) = 0, using ZFT

    x^2 = 0, that means a double root of zero, {0 ,0}and next is 4x^2 - 18 = 0,

    for 4x^2 - 18 = 0

    x^2 = 18/4

    x1 = (3/2) sqrt 2

    x2 = -(3/2) sqrt 2

    x = {0, 0, (3/2) sqrt 2, x2 = -(3/2) sqrt 2}

    - --------------------------------------------------------------------------------

    2),

    (3x + 1)^(1/3) = 5,

    raise to 3 both sides

    [(3x + 1)^(1/3)]^3 = [5]^3,

    3x + 1 = 125, solving for x, we have

    3x = 125 - 1,

    3x = 124

    x = 124/3 = 41.333333 . . . .

    CHECK: (3(124/3) + 1)^1/3 - 5 = 0,

    (124 + 1)^1/3 - 5 = 0

    125^1/3 - 5 = 0

    5 - 5 = 0

    0 = 0, indeed

    see graph
    Attached Thumbnails Attached Thumbnails Find all real solutions-fty.gif  
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  4. #4
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    Quote Originally Posted by pacman View Post
    1.

    2.

    Solution:

    1)

    Use factoring,

    x^2(4x^2 - 18) = 0, using ZFT

    x^2 = 0, that means a double root of zero, {0 ,0}and next is 4x^2 - 18 = 0,

    for 4x^2 - 18 = 0

    x^2 = 18/4

    x1 = (3/2) sqrt 2

    x2 = -(3/2) sqrt 2

    x = {0, 0, (3/2) sqrt 2, x2 = -(3/2) sqrt 2}

    - --------------------------------------------------------------------------------

    2),

    (3x + 1)^(1/3) = 5,

    raise to 3 both sides

    [(3x + 1)^(1/3)]^3 = [5]^3,

    3x + 1 = 125, solving for x, we have

    3x = 125 - 1,

    3x = 124

    x = 124/3 = 41.333333 . . . .

    CHECK: (3(124/3) + 1)^1/3 - 5 = 0,

    (124 + 1)^1/3 - 5 = 0

    125^1/3 - 5 = 0

    5 - 5 = 0

    0 = 0, indeed

    see graph
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  5. #5
    MHF Contributor
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    Quote Originally Posted by isundae View Post
    I was absent today, but managed to get the homework.
    1) Do the factorization of the common factor, moving this out front. Then factor the difference of squares that is left behind. Then set all three factors equal to zero, and solve each.

    2) To learn how to solve radical equations, try here. Your first step will be to add the 5 over to the other side of the equation.
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